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3 years ago

1) how does reflection differ from diffraction?

Physics

2 answers:

Evgesh-ka [11]3 years ago

5 0

Answer: (1) The correct answer is A.

(2) The correct answer is D.

Explanation:

(1)

Reflection is the sending back of light from the surface without absorbing it. In the reflection phenomenon, the wave does not continue moving forward.

Diffraction is the bending of the light around the obstacle. In the diffraction phenomenon, the wave travels forward after striking around the obstacle.

Therefore, the correct answer is A.

(2)

Amplitude is the maximum displacement in the medium from the rest position.

The amount of energy is related to the amplitude. Amplitude is related to the amount of energy carried by the wave. Low energy wave is characterized by a low amplitude. High energy wave is characterized by a high amplitude.

Therefore, the correct option is D.

ivann1987 [24]3 years ago

4 0

(1) Option (A) is correct. Reflection is the only process in which the wave does not continue moving forward.

(2) Option D are correct. Amplitude is related to the amount of energy carried by the wave as well as the maximum displacement from the rest position.

Further Explanation:

Part (1):

Reflection is defined as the phenomenon of bending of light in which the wave bounces back off the surface and changes its direction. Thus, does not continue to move further.

On the other hand, Diffraction is the process whose keen cause is the bending of the wave through small obstacles (which are comparable to wavelength of the wave) after which the wave always continues to move further.

Further, with reflection, the phenomenon like the diffraction and the interference are also defined by the change in the intensity of the wave that is observed on the screen for subsequent fringes. And the phenomenon of refraction also takes place in the mechanical waves (e.g. in the Rope of variable diameter, at the point of change in diameter both refraction and reflection takes place), also diffraction takes place in sound waves which are also mechanical waves.

Lastly, with reflection, refraction is also defined by the change in the direction of the light ray as they either bend towards or away from the normal line as the light rays enters into the different mediums.

Part (2):

The amplitude is related to the amount of the energy carried by the wave and is given as:

$\boxed{E=kA^2}$

Here, $E$ is the energy carried by the wave, $k$ is the wave number and $A$ is the amplitude of the wave.

Further, the amplitude is defined as the maximum magnitude of the oscillating physical quantity associated with the wave. (e.g. If a particle is oscillating then its displacement is the function of time and is a repetitive variable, so the maximum value of displacement is called as amplitude.)

Therefore, option D are correct. Amplitude is related to the amount of energy carried by the wave as well as the maximum displacement from the rest position.

Learn More:

1. What type of mirror is used by the dentist brainly.com/question/997618

2. A microwave oven operates at 2.70 Ghz brainly.com/question/9077368

Answer Details:

Grade: High School

Subject: Physics

Chapter: Optics

Keywords:

Reflection, refraction, diffraction, interference, light, light ray, light wave, mechanical waves, sound, bending of light, obstacle, amplitude, maximum displacement, rest, energy carried by the wave, change in the direction.

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A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the

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Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.

(a) Find the vertical component of the velocity with which the ball hits the ground.

(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

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